If $\int \frac{\sin x - \cos x}{(\sin x + \cos x)\sqrt{\sin x\cos x + \sin^2 x\cos^2 x}}dx = \cos ec^{-1}(g(x)) + cY \in \mathbb{R}$, then
Step-by-Step Solution
Key Concept: Substitute $u = \sin x + \cos x$ to convert the integral into inverse cosecant form, identifying $g(x) = 1 + \sin 2x$.
Let $u = \sin x + \cos x$, so $du = (\cos x - \sin x)dx$. The denominator becomes $u\sqrt{\sin x\cos x + \sin^4 x\cos^2 x}$. Note that $\sin 2x = 2\sin x\cos x$, so $\sin x\cos x = \frac{\sin 2x}{2}$. The integral becomes $-\int \frac{du}{u\sqrt{\frac{\sin 2x}{2} + \frac{\sin^2 2x}{16}}}$. Factoring out from the square root and simplifying yields $\csc^{-1}(1 + \sin 2x)$. Therefore $g(x) = 1 + \sin 2x$. Since $-1 \leq \sin 2x \leq 1$, we have $0 \leq g(x) \leq 2$, so $g(x) \geq 0$.
Correct Answer: 1,3